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Derivation of Continuity Equation explains the law of conservation of mass in fluid dynamics. Also, if the fluid is incompressible, the density will remain constant for steady flow. So, ρ1 =ρ2. It is a formulation of the law of mass conservation. The term fluid is commonly used to describe a liquid or an incompressible fluid, but it can also be used to describe a gas. Fluid dynamics describes patterns of fluid flow. There are two major ways that fluids are forced to flow. Fluids can flow downhill owing to gravity, or fluids can flow due to pressure differences.
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Key Terms: Continuity Equation, Fluid Dynamics, Mass Conservation, Gravity
Continuity Equation
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According to the continuity equation, the product of the cross-sectional area of the pipe and the fluid speed at any point along the pipe is always constant. The volume flow per second, or flow rate, is equivalent to the product of the cross-sectional area of the pipe and the fluid speed . The equation for continuity is as follows:
R = A v = constant
Where,
R is the volume flow rate
A is the flow area
v is the flow velocity
Assumption of Continuity Equation
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Based on these assumptions, the continuity equation is derived:
- Tubes have only one entry and one exit
- Fluid flowing through the tube has no viscosity
- The flow is incompressible
- The fluid flow is steady
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Derivation of Continuity Equation
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Consider that the fluid flows in the tube for a short period of time. Assume that Δt is a small period of time. At the bottom end of the pipe, the fluid will traverse a distance of Δx1 at a velocity of v1.
The fluid will cover the following distance at this time:
Δx1 = v1Δt
The volume of water that will flow into the pipe at the lower end is:
V = A1 Δx1 = A1*v1*Δt
We know, mass (m) = Density ρ × Volume V.
Hence, in the Δx1region, the fluid mass will be:
Δm1= Density × Volume
=> Δm1 = ρ1A1v1Δt ...Eq.1
At this point, the mass flux must be estimated at the lowest level. The mass of the fluid per unit time moving through any cross-sectional region is simply described as mass flux. The mass flux at the lower end, with cross-sectional area A1, will be:
Δm1/Δt = ρ1A1v1 ...Eq.2
At the upper end, mass flux is as follows:
Δm2/Δt = ρ2A2v2 ...Eq.3
Here, v2 is the fluid velocity through the upper end of the pipe, i.e. via Δx2, in Δt time, and A2 is its cross-sectional area.
As long as the flow is continuous, the density of the fluid between the lower and higher ends of the pipe remains constant throughout time.
As a result, both mass fluxes at the lower end and the upper end of the pipe are equal. Equation 2 = Equation 3.
Thus,
ρ1A1v1 = ρ2A2v2 ...Eq.4
This can also be written as:
ρ A v = constant
In fluid dynamics, the equation establishes the law of mass conservation. Also, with continuous flow, if the fluid is incompressible, the density will stay constant. Therefore, ρ1 = ρ2.
Therefore, Equation 4 is now written as follows:
A1 v1 = A2 v2
In general form, this equation is as follows:
A v = constant
Taking R as the volume flow rate, the above equation becomes:
R = A v = constant
Applications of the Equation of Continuity
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- Hydrodynamics, Electromagnetism, Aerodynamics, and Quantum Mechanics are the most common applications of the Equation of Continuity.
- The essential rule of Bernoulli's Principle is based on the equation of continuity.
- It is also linked to the principle of aerodynamics and its applications.
- The consistency of Maxwell's Equation is determined using the differential form of the equation of continuity. In addition, in electromagnetism, the differential form of the equation of continuity is utilised.
- The Schrodinger Equation is checked for consistency using the continuity equation.
- The equation of continuity is also used in the Theory of Relativity and Noether's Theorem.
Things to Remember Based on Derivation of Continuity Equation
- The continuity equation asserts that in a steady flow, the quantity of fluid flowing through one point must be equal to the amount of fluid flowing through another point, or the mass flow rate must be constant.
- It is another formulation of the law of mass conservation.
- The continuity equation is given by R = Av = constant. Where, R is the volume flow rate, A is the flow area and v is the flow velocity
- This equation is linked to the principle of aerodynamics and its applications.
- The differential form of the equation of continuity is used in electromagnetism.
Important Question Based on Derivation of Continuity Equation
Ques: A liquid flows at a speed of 9cm/s through a pipe with a diameter of 10cm. What is the new velocity of the liquid if the pipe diameter is reduced to 6cm? (3 Marks)
Ans: The rate of flow, A * v, must remain constant.
By using the continuity equation, A1v1=A2v2.
Solving the initial cross-sectional area we get,
A1=πr2 = 25πcm2.
The initial radius is 5cm.
Then find the final area of the pipe:
A2=πr2 = 9πcm2.
The final radius is 3cm.
Using these values in the continuity equation allows us to solve the final velocity.
(25πcm2)(9cm/s)=(9πcm2)v2
v2=25cm/s
Ques: If blood flows through the aorta with velocity, va, with what velocity would blood flow through the capillaries in the body? (3 Marks)
Ans: The flow rate of blood through the body is equal to area times velocity, just as the volume flow rate equation for fluids.
Flow rate=Av
The flow rate is constant, therefore the velocity varies depending on the region through which the blood passes; hence, the volume flow rate via the aorta is equal to the volume flow rate through the capillaries.
Aava=Acvc
Because the area of the aorta and the area of the capillaries can be determined, the velocity through the aorta may be used to calculate the velocity through the capillaries.
vc=Aava/Ac
Ques: What is the relationship between flow speed at point 2 and point 1 if a pipe with flowing water has a cross-sectional area nine times bigger at point 2 than at point 1? (3 Marks)
Ans: Using the continuity equation we know that A1V1=A2V2.
The question tells us that the cross-sectional area at point 2 is nine times greater than at point 1 9A1=A2
Using the continuity equation, A1= 1 and A2 = 9.
1V1=9V2
V1/V2=9/1
At point 1, the flow speed is nine times greater than at point 2.
Ques: Blood moves at a velocity of v via an artery. What is the new blood velocity if a vasoconstricting substance is eaten and the artery constricts to half its original diameter? (4 Marks)
Ans: The continuity equation states that:
A1v1=A2v2
To put it another way, the volumetric flow rate remains constant in a pipe of increasing diameter. The velocity must rise as the diameter shrinks.
We need to find the area in terms of the diameter to get the change in cross-sectional area:
A = πr2 = π(D/2)2 = πD2/4
When the diameter is halved, the area is quartered.
A2 = π (D/2)2 / 4 = π (D2/4) /4 = ¼ * π D2 /4
A2=¼ A
Increasing the velocity by a factor of four is necessary to maintain a constant volumetric flow.
A1v1=(¼ A1)v2
v2=4v1
Ques: A garden hose with a nozzle is connected to a pipe with a diameter of 4 cm. What is the velocity of the flow at the nozzle when it is modified to have a diameter of 8 millimetres if the velocity of flow in the pipe is 2ms? (4 Marks)
Ans: In order to generate linear flow, the flow rate in a pipe must be constant. The product of the cross-sectional area and the fluid velocity determines the flow rate.
A1v1 = A2v2
The radii of the pipe and nozzle may be used to determine their cross-sectional areas. Remember that the dimensions were given in terms of diameter, so divide by 2 to get the radius.
A=πr2
A1 = π(0.02m)2 = 0.0004πm2
A2 = π(0.004m)2 = 0.000016πm2
Calculate the final velocity in the nozzle using these areas and the initial velocity.
(0.0004πm2)(2m/s) = (0.000016πm2)v2
v2=(0.0004πm2)(2m/s)0.000016πm2
v2=50m/s
Ques: On a circular conduit, there are different diameters: diameter D1 = 2 m changes into D2 = 3 m. The velocity in the entrance profile was measured: v1 = 3 ms-1. Calculate the discharge and mean velocity at the outlet profile (see figure below). Determine also type of flow in both conduit profiles (whether the flow is laminar or turbulent) – temperature of water T = 12° C. (4 Marks)
Ans: Discharge Q and consequently velocity v2 can be calculated from the continuity equation
To determine the type of flow in a conduit, the Reynolds number \(Re = \frac{v.D}{v}\) will be used. For laminar flow: Re<2320 For turbulent flow Re>2320 Kinematic viscosity of water of 12°C: v= 124.10-6 m2.s-1 (see Tab. 1)
For the conduit:
Ques: A water clock is an axisymmetric vessel with a small exit pipe at the bottom. Find the shape for which the water level falls at equal heights in equal intervals of time. (5 Marks)
Ans: Continuity equation: \(\frac{d}{dt} (pv) + m_{out} - m_{in} = 0\)
There is no inflow and the fluid is incompressible. Let ‘a’ be the cross-section of the exit pipe:
Bernoulli equation between the upper free surface and the exit section:
Thus, using the first relation:
From the question, we know that the variation of h is linear in time:
h(t) = h0 – xt
Thus,
The volume of fluid can always be written as:
Ques: A horizontal pipe with a constriction is called a Venturi Tube and is used to measure flow velocities by measuring the pressure at two different cross-sectional areas of the pipe. Given two pressures P1 and P2 where the areas are A2 and A1 respectively, determine the flow velocity at point 2 in terms of these quantities and the fluid density ρ. (2 Marks)
Ans: First, use Bernoulli’s law, and take the heights y1 = y2 = 0:
Now substitute for one of the velocities, v1, by using the continuity equation:
Ques: A large storage tank filled with water develops a small hole in its side at a point 16 m below the water level. If the rate of flow from the leak is 2.5 x 10−3 m3/min, determine:
a) the speed at which the water leaves the hole
b) the diameter of the hole (4 Marks)
Ans: We assume that the tank and the hole are both open to the atmosphere. Call the top position 1 and the point of the hole position 2. So P1 = P2 = Pa. We now write Bernoulli’s law:
The continuity equation allows us to relate the speeds to the areas at the two positions
\(v_1 A_1 = V_2 A_2 \implies V_1 = \frac{A_2}{A_1} V_2\)
Because the area A1 \(\gg\) A2 we can ignore v1 in comparison with v2 (v1 \(\ll\) v2) Now substitute v1 = 0 and cancel out the equal pressures in Bernoulli’s law to get:
For part b) we know that the volume flow rate is the product of the area of the hole and the velocity
flow rate = Av
We first convert the flow rate given in m3/minute into m3/second by dividing by 60. This gives 4.167 x 10−5 m3/second
4.167 x 10−5 = A2v2 = A2·17.7 = ⇒ A2 = .2354 x 10−6 m2
This is equivalent to a diameter of 0.0017 meters.
Ques: A pipe with a diameter of 4 centimeters is attached to a garden hose with a nozzle. If the velocity of flow in the pipe is 2ms, what is the velocity of the flow at the nozzle when it is adjusted to have a diameter of 8 millimeters? (4 Marks)
Ans: The flow rate in a pipe must be constant in order to create a linear flow. This flow rate is given by the product of the cross-sectional area and the velocity of the fluid.
A1v1=A2v2
The cross-sectional areas of the pipe and nozzle can be found using their radii. Note that you were given dimensions in terms of diameter, so be sure to divide by 2 to get the radius.
A=πr2
A1=π(0.02m)2=0.0004πm2
A2=π(0.004m)2=0.000016πm2
Use these areas and the initial velocity to calculate the final velocity in the nozzle.
(0.0004πm2)(2ms)=(0.000016πm2)v2
v2=(0.0004πm2)(2ms)0.000016πm2
v2=50ms
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